A 74.0 kg ice skater moving to the right with a velocity of 2.58 m/s throws a 0.18 kg snowball to the right with a velocity of 24.4 m/s relative to the ground.
(a) What is the velocity of the ice skater after throwing the snowball? Disregard the friction between the skates and the ice.
(b) A second skater initially at rest with a mass of 60.00 kg catches the snowball. What is the velocity of the second skater after catching the snowball in a perfectly inelastic collision?
To solve this problem, we can use the principle of conservation of momentum. The total momentum before the snowball is thrown is equal to the total momentum after the snowball is thrown.
(a) To find the velocity of the ice skater after throwing the snowball, we can use the equation:
(m1 * v1) + (m2 * v2) = (m1 * u1) + (m2 * u2)
Where:
m1 = mass of the ice skater
v1 = velocity of the ice skater before throwing the snowball
m2 = mass of the snowball
v2 = velocity of the snowball before being thrown
u1 = velocity of the ice skater after throwing the snowball (unknown)
u2 = velocity of the snowball after being thrown (which is relative to the ground)
Given:
m1 = 74.0 kg
v1 = 2.58 m/s
m2 = 0.18 kg
v2 = 24.4 m/s
Plugging in these values into the equation, we get:
(74.0 kg * 2.58 m/s) + (0.18 kg * 24.4 m/s) = (74.0 kg * u1) + (0.18 kg * 24.4 m/s)
Simplifying the equation:
190.92 kg m/s + 4.45 kg m/s = 74.0 kg * u1 + 4.39 kg m/s
Rearranging the equation to solve for u1:
74.0 kg * u1 = 190.92 kg m/s + 4.45 kg m/s - 4.39 kg m/s
74.0 kg * u1 = 191.98 kg m/s
Dividing both sides by 74.0 kg:
u1 = 191.98 kg m/s / 74.0 kg
u1 ≈ 2.59 m/s
Therefore, the velocity of the ice skater after throwing the snowball is approximately 2.59 m/s to the right.
(b) In a perfectly inelastic collision, the two objects stick together and move with a common velocity after the collision. Let's use the conservation of momentum to find the velocity of the second skater after catching the snowball.
(m1 * v1) + (m2 * v2) = (m1 + m2) * v
Where:
m1 = mass of the snowball
v1 = velocity of the snowball before the collision (relative to the ground)
m2 = mass of the second skater
v2 = velocity of the second skater before the collision
(v is the common velocity after the collision)
Given:
m1 = 0.18 kg
v1 = 24.4 m/s
m2 = 60.0 kg
v2 = 0 m/s (since the second skater is initially at rest)
Plugging in these values into the equation, we get:
(0.18 kg * 24.4 m/s) + (60.0 kg * 0 m/s) = (0.18 kg + 60.0 kg) * v
Simplifying the equation:
4.392 kg m/s = 60.18 kg * v
Rearranging the equation to solve for v:
60.18 kg * v = 4.392 kg m/s
Dividing both sides by 60.18 kg:
v = 4.392 kg m/s / 60.18 kg
v ≈ 0.073 m/s
Therefore, the velocity of the second skater after catching the snowball in a perfectly inelastic collision is approximately 0.073 m/s to the right.